# Separation of variables..

• May 4th 2009, 04:24 AM
Raidan
Separation of variables..
I get stuck at the integration step(Wondering)

use the method of separation of variables to find the solution of the differential equation:

dy/dx=sqrt(2y-y^2)/x+1 , y(0)=1
• May 4th 2009, 04:44 AM
Jester
Quote:

Originally Posted by Raidan
I get stuck at the integration step(Wondering)

use the method of separation of variables to find the solution of the differential equation:

dy/dx=sqrt(2y-y^2)/x+1 , y(0)=1

So $\int \frac{dy}{\sqrt{2y-y^2}} = \int \frac{dx}{x+1}$
then complete the square in the square root term

$
\sqrt{2y-y^2} = \sqrt{1-(y-1)^2}
$
Let $u = y-1$ and you should be able to integrate.
• May 4th 2009, 06:30 AM
Raidan
could someone help me to solve this question?(Bow)
• May 4th 2009, 06:36 AM
e^(i*pi)
Quote:

Originally Posted by Raidan
could someone help me to solve this question?(Bow)

Danny pointed it out:

u = y-1 so du = dy

$\int{\frac{du}{\sqrt{1-u^2}}} = \int{\frac{dx}{x+1}}$
The left side is a standard integral and is equal to $arcsin(u) +A$
For the right side recall that $\int{\frac{f'(x)}{f(x)}} = ln|f(x)| + B$